Tính hợp lý tổng sau:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
Ai nhanh mk tick, có cả cách làm và lời giải nữa nhé!
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
ai trả lời nhanh nhất và đúng nhất sẽ đc tick^-^!!!!!!!!!!!!!!!!!!!!!!!
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\left(1-\frac{1}{50}\right)+\left(\frac{-1}{2}+\frac{1}{2}+\frac{-1}{3}+\frac{1}{3}+...+\frac{1}{49}\right)\)
\(A=\frac{49}{50}+0\)
\(A=\frac{49}{50}\)
Tính hợp lý tổng sau :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=\frac{1}{1}-\frac{1}{50}\)
\(A=\frac{50-1}{50}=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\\ =1-\frac{1}{50}=\frac{49}{50}\)
Chứng minh rằng:
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+....+\frac{1}{50}\)
Ai làm nhanh, chi tiết thì mk tick cho nhé!!!
Vế trái:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
=\(\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
=\(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)=Vế phải
Tính hợp lý
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
Giúp mình với!!!!Thanks =)))
1/1.2+1/2.3+1/3.4+.....+1/49.50
=1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50
=1-1/50
=49/50
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Tính tổng sau một cách hợp lí :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}+\frac{1}{2018.2019}\)
Làm được t tick
gọi biểu thức trên là A A=1/1 -1/2+1/3-1/4+...+1/2017-12018+1/2018-1/2019 A=1/1-1/2019 A=2018/2019
1/1.2+1/2.3+1/3.4+1/4.5+...+1/2017.2018+1/2018.2019
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
cái ĐỒ ĐÁNG GHÉT ◥ὦɧ◤ŤŔầŃ VăŃ ĤùŃĞ™ kia t định trả lời sao m dám....
Tính tổng: A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
= 1-\(\frac{1}{50}\)
= \(\frac{49}{50}\)
ta có công thức tính tổng quát 1/[n(n+1)] = 1/n -1/(n+1)
=> A=1/1.2+ 1/2.3+1/3.4+1/4.5+...+1/49.50
=1/1 -1/2 +1/2 -1/3 +1/3-1/4+.......+1/49 -1/50
= 1 -1/50 = 49/50
Ai thấy đúng thì tk cho mk nhé
= \(\frac{49}{50}\).
Đúng 100% luôn!
Chúc các bạn học giỏi.
Tính nhanhB=\(-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-.....-\frac{1}{98.99}-\frac{1}{99.100}\)
Giups mình với mình đang cần gấp ai nhanh nhất tớ sẽ tích cho bạn đó có kèm cách làm nhé
\(B=-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\\
=-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\\
=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\\
=-\left(1-\frac{1}{100}\right)=\frac{-99}{100}\)
<br class="Apple-interchange-newline"><div></div>B=−11.2 −12.3 −13.4 −...−198.99 −199.100 \\ =−(11.2 +12.3 +13.4 +...+198.99 +199.100 ) \\ =−(1−12 +12 −13 +13 −14 +...+198 −199 +199 −1100 ) \\ =−(1−1100 )=−99100
Tính một cách hợp lí tổng sau :
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}.\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
Tìm x
\(X-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{98.99}=\frac{1}{100}+\)1/99.100
Mình đang cần gấp giúp mình với ai làm đúng tớ sẽ tích cho bạn đó, nêu cả cách giải giúp mình nhé!
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)